Prime Octet 2

It's evident that $\color{#EC7300}B \color{#333333}\rightarrow \color{#69047E}H$ are not primes, because we can always factor out certain numbers i.e

$\color{#EC7300}B \color{#333333} = 16! + 2 = 2( 3 \cdot 4 \cdot...\cdot 15 \cdot 16 +1)$

$\color{gold}C \color{#333333} = 16! + 3 = 3( 2 \cdot 4 \cdot...\cdot 15 \cdot 16 +1)$

$\vdots$

$\color{#69047E}H \color{#333333} = 16! + 8 = 8( 2 \cdot 3 \cdot ... \cdot 7 \cdot 9 \cdot ... \cdot 15 \cdot 16 +1)$

$\color{#D61F06}A \color{#333333} = 16! + 1$ is quite an exception. We can utilise Wilson's Theorem.

Wilson's Theorem states that

For a positive integer $n > 1$ , if $(n-1)!\equiv -1\pmod n,$ then $n$ is a prime.

We use $n = 17$ which gives us:

$(16)! \equiv -1 \pmod {17}$

This means that:

$\color{#D61F06}A \color{#333333} = 16! + 1$ is divisible by $17$

Therefore, $\color{#D61F06}A \color{#333333} = 16! + 1$ is not a prime. Hence, all eight numbers above are not primes.